Hydrocarbons
Treatment of buta-1,3-diyne with NaNH2 (2 equivalents), followed by reaction with excess of trans-CH3-CH=CH-CH2-Br gives X as the major product. The maximum number of carbon atoms that are collinear (in a straight line) in X is _____.
Explanation:
Consider the depicted hydrogen (H) in the hydrocarbons given below. The most acidic hydrogen (H) is
B forms most stable conjugate base and hence $B$ is most acidic$
\text { Answer: Option (B) }
$
In the following reaction sequence, the major product $\mathbf{P}$ is formed.
Glycerol reacts completely with excess $\mathbf{P}$ in the presence of an acid catalyst to form $\mathbf{Q}$. Reaction of $\mathbf{Q}$ with excess $\mathrm{NaOH}$ followed by the treatment with $\mathrm{CaCl}_2$ yields Ca-soap $\mathbf{R}$, quantitatively. Starting with one mole of $\mathbf{Q}$, the amount of $\mathbf{R}$ produced in gram is __________.
[Given, atomic weight: $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{Cl}=35, \mathrm{Ca}=40$ ]
Explanation:
$\therefore 1$ mole $Q$ produce $\frac{3}{2}$ moles $R$
$\therefore$ Mass of $R$ produce $=\frac{3}{2} \times 606=909 \mathrm{~g}$
Explanation:
Explanation:
Explanation:
If the reaction sequence given below is carried out with 15 moles of acetylene, the amount of the product $\mathbf{D}$ formed (in $\mathrm{g}$ ) is ___________ .
The yields of $\mathbf{A}, \mathbf{B}, \mathbf{C}$ and $\mathbf{D}$ are given in parentheses.
[Given: Atomic mass of $\mathrm{H}=1, \mathrm{C}=12, \mathrm{O}=16, \mathrm{Cl}=35$ ]
Explanation:
Molecular formula of D is C8H8O2
Molar mass of D is ($12\times8+8\times1+16\times2$) = 136 g
$\therefore$ Mass of D is 136
x g and y g are mass of R and U, respectively.
(Use : Molar mass (in g mol$-$1) of H, C and O as 1, 12 and 16, respectively)
The value of x is ________.
Explanation:
$M{g_2}{C_3} + 4{H_2}O \to 2Mg{(OH)_2} + C{H_3} - \mathop {\mathop C\limits_{(P)} }\limits_{{\mathop{\rm Propyne}\nolimits} } \equiv CH$
Sodium amide (NaNH2) takes proton from alkyne to form acetylide ion which, attacks on Me$-$I to give an alkyne with one more carbon.
Mass of P formed = 4.0 g
Molar mass of P formed = 40 g mol$-$1
Number of moles of P = ${{Mass\,of\,P(W)} \over {Molar\,mass\,of\,P(M)}} = {4 \over {40}} = 0.1$ mol
1 mole of P forms 75% i.e., 0.75 = 0.075 mole of Q
Polymerisation of alkyne (Butyne) occurs in presence of red hot iron tube to form benzene derivative.
3 moles of alkyne forms 40% i.e., 0.40 mole of benzene derivative, R.
0.075 mole of Q forms ${{0.40} \over 3} \times 0.075 = 0.01$ mole of R.
Molar mass of R = 162 g/mol
Mass of R formed or value of x = Molar mass $\times$ Number of moles = 162 $\times$ 0.01 = 1.62 g
Explanation:
$ 60+32+8=100 $
The value of $\mathrm{Y}=0.032 \times 100=3.2$
Choose the correct option(s).
Explanation:
Compound (P) has total number of hydroxyl groups = 6
Among the following reaction(s), which gives(give) tert-butyl benzene as the major product is(are)
In the following reactions, the product S is
The major product U in the following reactions is
In the following reaction, the major product is
The correct statement with respect to product Y is
Among P, Q, R and S, the aromatic compound(s) is(are)
The major product of the following reaction is
The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is _______.
Explanation:
(2) Since, bromide (Br–) is a good leaving group, elimination using strong base takes place using strong base (KOH) take via E2 mechanism. This is also called $\beta $ elimination.
(3) There are 3 different types of protons :
(iv) The strong base abstracts $\beta $ hydrogen (H1 or H2 or H3) with simultaneous loss of bromide ion forming an alkene. This alkene can exist in 2 conformations, E and Z.
(a) Elimination of H1 proton :
This product can exist in 2 conformations, E and Z.
(b) Elimination of H2 proton :
Since, the groups about sp2 hybridised carbon of cyclopentane is fixed no E and Z forms are possible. This molecule has only one conformation.
(c) Elimination of H3 proton :
This product can exist in 2 conformations, E and Z.
Hence, a total of 5 products are possible.
The maximum number of isomers (including stereoisomers) that are possible on mono-chlorination of the following compound, is ____________.
Explanation:
(i) The monochlorination products obtained depends upon the kinds of hydrogen present in the molecule.
(ii) There are four different types of hydrogen in the molecule; $\mathrm{H}_a, \mathrm{H}_b, \mathrm{H}_c$ and $\mathrm{H}_d$.
(iii) The monochlorination products obtained on replacement of these hydrogen are as follows :
(a) Replacing $\mathrm{H}_a$ with chlorine :
Since, the carbon is a chiral; only one monochlorination isomer is possible when $\mathrm{H}_a$ is replaced by chlorine.
(b) Replacing $\mathrm{H}_b$ with chlorine :
Since all the carbons are a chiral; only one monochlorination product is possible by when $\mathrm{H}_b$ replaced by Cl .
(c) Replacing $\mathrm{H}_c$ with chlorine
The carbon on which hydrogen is replaced by chlorine becomes chiral; hence R/S isomers exist.
Also, carbon adjacent to the carbon on which hydrogen $\left(\mathrm{H}_c\right)$ is replaced, is also chiral; hence, it will also exist as a pair of enantiomer $R / S$ four monochlorination products (which are stereoisomers or optical isomers) are possible.
(d) Replacing $\mathrm{H}^d$ by chlorine :
The central carbon $\left({ }^*\right)$ becomes chiral when $\mathrm{H}^d$ is replaced by chlorine; hence $R / S$ isomers exist. Two monochlorination products are possible (stereoisomers or optical isomers) are possible.
Total number of isomers possible are 8.
In the following reaction,
The structure of the major 'X' is
The reagent(s) for the following conversion,
$ \begin{aligned} &\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{NOCl} \rightarrow \mathrm{P}\\ &\text { Identify the adduct. } \end{aligned} $
Step 3:
The IUPAC name of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}$ is
Benzoyl chloride
Benzene chloroketone
Benzene carbonyl chloride
Chlorophenyl ketone
The IUPAC name of the compound is benzene carbonyl chloride.
$ \text { The major products } \mathbf{P} \text { and } \mathbf{Q} \text { are } $
$ \text { It is a two step reaction: } $
Hence, product $P$ is cumene.
The cumene hydroperoxide on acidic hydrolysis produces phenol and acetone.
Explanation:
Now, assume moles of ${C_2}{H_6} = x$
$\therefore$ Moles of ${C_2}{H_5}Br = x$ when 100% yield.
Given that 90% yield of ${C_2}{H_5}Br$ happens.
$\therefore$ Moles of ${C_2}{H_5}Br = x \times {{90} \over {100}} = 0.9x$
Now, from 2 moles of ${C_2}{H_5}Br$ 1 mole of ${C_4}{H_{10}}$ produced.
$\therefore$ From 0.9x moles of ${C_2}{H_5}Br$ ${1 \over 2} \times 0.9x$ moles of ${C_4}{H_{10}}$ produced when 100% yield happens. But given that 85% yield of ${C_4}{H_{10}}$ happens.
$\therefore$ Moles of ${C_4}{H_{10}} = {1 \over 2} \times 0.9x \times {{85} \over {100}}$
$ = {{(0.9 \times 0.85)x} \over 2}$
According to question,
${{(0.9 \times 0.85)x} \over 2} = {{55} \over {58}}$
$ \Rightarrow x = {{55} \over {29 \times 0.9 \times 0.85}} = 2.48$
$\therefore$ Volume of ${C_2}{H_6} = 2.48 \times 22.4$ Litres
$ = 55.552$ Litres